[Pb2+]
= (1.1x10^-4 mol/L x 0.02L)/(0.02L + 0.08L)
= 2.2x10^-5 mol/L
This is right for Pb^2+.
[I-]
= (4.45x10^-2 mol/L x 0.08L)/0.1L
= 3.56x10^-2 mol/L
This is right for [CaI2] but [I^-] is twice that or 7.12E-2
Qsp = [Pb2+][I-]^2
= (2.2x10^-5)(3.56x10^-2)^2
= 2.7x10^-8 This is your Qsp but it isn't right since I^- is wrong. It is not Ksp
That makes Qsp = (2.2E-5)(7.12E-2)^2 = 1.15E-7 which is what you compare with Ksp. I don't know what Ksp value you are using but you should use the one in your text/notes. The value I found on the web was approx 8E-9 so a ppt should form since Qsp > Ksp.
20.00mL of a 1.100x 10^-4 mol/L Pb(NO3)2 is mixed with 80.00mL of 4.45010^-2 mol/L CaI2. Will a precipitate form?
for the Ksp i got 2.7x10^-8 and im not sure if its right or not
this is what i did
[Pb2+]
= (1.1x10^-4 mol/L x 0.02L)/(0.02L + 0.08L)
= 2.2x10^-5 mol/L
[I-]
= (4.45x10^-2 mol/L x 0.08L)/0.1L
= 3.56x10^-2 mol/L
Qsp = [Pb2+][I-]^2
= (2.2x10^-5)(3.56x10^-2)^2
= 2.7x10^-8
did i do anything wrong??
1 answer
2 answers